Hyperbolic type channel MOSFET

ABSTRACT

In a MOSFET, a source region, a drain region and a channel region disposed between the source region and the drain region are provided. And the width W(x) of the channel region is changed according to the following mathematical equation.          W        (   x   )       =                 4          {         (     1   +   δ     )          E   sat        L     +     V   DD     -     V   T       }     2       +                 {         (     1   +   δ     )          E   sat        L     +     2        (       V   DD     -     V   T       )         }          (       V   DD     -     V   T       )               6        {         (     1   +   δ     )          E   sat          L        (     1   -     x   L       )         +     V   DD     -     V   T       }          {         (     1   +   δ     )          E   sat        L     +     V   DD     -     V   T       }              W   O                       
     In the equation, x denotes a distance from the end of the source region side to the direction toward the drain region side, W(x) denotes the width of the channel region at the position of a distance x, δ denotes a substrate charge effect coefficient, L denotes the length from the end of the source region side to the drain side, V DS  denotes a gate voltage, V T  denotes a threshold voltage, W O  denotes a standard width of the channel region, and E sat  denotes the strength of the carrier velocity saturation electric field.

BACKGROUND OF THE INVENTION

1. Field of the Invention

The present invention relates to a MOSFET (metal oxide semiconductor field effect transistor) in a semiconductor device and, in particular, to a hyperbolic type channel MOSFET capable of increasing a drain current by determining the width of a channel region according to distances from a source region and a drain region.

2. Description of the Related Art

In recent years, the packaging density of the semiconductor device is increasing and hence a further fine patterning has been required also in the MOSFET. For this reason, the drain current density in the MOSFET has been required to increase.

In order to increase a drain current, there has been conventionally used a method of utilizing the overshoot effect of a carrier. The overshoot effect of a carrier is produced by applying a strong electric field to a channel region of a MOSFET and by shortening the length of the channel region from a source region to a drain region (hereinafter referred to as “channel region length”).

In a conventional MOSFET, however, there is presented a problem that even if the electric field applied to the channel region is made stronger, a current is not so much increased. FIG. 1 is a plan view to show the configuration of a conventional MOSFET. As shown in FIG. 1, the conventional MOSFET is composed of a source region 11, a drain region 12, and a channel region 13 disposed between the source region 11 and the drain region 12. The channel region 13 is rectangular. That is, the width of the channel region 13 is set at a constant value W₀ along the entire length from the end of the source region 11 side to the end of the drain region 12 side. Further, the width of the source 11 region and the width of the drain region 12 are equal to the width of the neighboring channel region 13. Still further, an insulating layer 14 is formed on the channel region 13 (at the front side of the paper in FIG. 1) and a gate 15 is formed on the insulating layer 14.

In the channel region 13 of the MOSFET shown in FIG. 1, as a position comes near to the end of the drain region 12 side from the end of the source region 11 side along the channel, a difference between a channel potential and a gate potential decreases and the amount of carriers induced in an inversion layer decreases. Since a current continuity law is established in the direction of channel, in order to realize the drain current of a predetermined magnitude to compensate the decreased amount of carriers, a stronger electric field is required to be applied to the channel region to drift the carriers. However, as described above, the amount of carriers in the channel region 13 decreases as the position comes near to the drain region 12 and hence the channel potential increases in gradient as the position comes near to the drain region 12, and hence the potential difference between the source region 11 and the drain region 12 is applied mainly to the vicinity of the drain side end of the channel region 13.

FIG. 2 is a graph to show a channel potential distribution in the conventional MOSFET, in which a horizontal axis designates a distance x from the end of the source region 11 side of the channel region 13 to the direction toward the drain region 12 side and a vertical axis designates a channel potential Vcs₀. As shown in FIG. 2, since a very strong electric field is not applied in the direction of the channel in the source region 11 side of the channel region 13, even if the MOSFET has a short channel length, it can not sufficiently benefit the velocity overshoot effect of the carriers because it is hard to produce the velocity overshoot effect of the carriers. As a result, the conventional MOSFET presents a problem that even if the channel length is made short, a drain current value is not improved so much.

SUMMARY OF THE INVENTION

An object of the present invention is to provide a MOSFET capable of making the strength of the electric field uniform in the direction of a channel and increasing a drain current.

A MOSFET according to the present invention comprises: a source region; a drain region; and a channel region disposed between the source region and the drain region. In the channel region, the width of the channel region changes according to the following mathematical equation (1). $\begin{matrix} {{W(x)} = {\frac{\begin{matrix} {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}^{2}} +} \\ {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{DD} - V_{T}} \right)}} \right\} \left( {V_{DD} - V_{T}} \right)} \end{matrix}}{6\left\{ {{\left( {1 + \delta} \right)E_{sat}{L\left( {1 - \frac{x}{L}} \right)}} + V_{DD} - V_{T}} \right\} \left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}}W_{O}}} & (1) \end{matrix}$

In equation (1), x denotes a distance from the end of the source region side to the direction toward the drain region side, W(x) denotes the width of the channel region at the position of the distance x, δ denotes a substrate charge effect coefficient, L denotes the length from the end of the source region side to the end of the drain side, V_(DD) denotes a gate voltage, V_(T) denotes a threshold voltage, W_(O) denotes a standard width of the channel region, and E_(sat) denotes a strength of a carrier velocity saturation electric field.

In the present invention, by setting the width of the channel region at the value described above, the total amount of carriers in the direction of width at a position in the channel is not dependent on a distance from the source region and the drain region but is made constant, so an electric field of the same strength can be applied to the entire channel region to maximize a drain current.

Further, it is preferably that, in the above-mentioned MOSFET, the width of the source region is equal to or larger than the width of the channel region at the source region side and the width of the drain region is equal to or larger than the width of the channel region at the end of the drain region side. This can produce the effect described above to a maximum level.

BRIEF DESCRIPTION OF THE DRAWINGS

In the accompanying drawings:

FIG. 1 is a plan view to show the configuration of a conventional MOSFET;

FIG. 2 is a graph to show a channel potential distribution in a conventional MOSFET, in which a horizontal axis designates a distance x from the end of the source region 11 side of the channel region 13 to the direction toward the drain region 12 side and a vertical axis designates a channel potential Vcs₀;

FIG. 3 is a plan view to show the configuration of a MOSFET according to the embodiment of the present invention;

FIG. 4 is a graph to show a relationship between x and W(x) satisfying the mathematical equation (2), in which a horizontal axis designates a distance x from the end of the source region 1 side of the channel region 3 to the direction toward the drain region 2 side and a vertical axis designates a width W(x) of the channel region 3 at the position of a distance x;

FIG. 5 is a graph to show a channel potential distribution of the MOSFET according to the present embodiment, in which a horizontal axis designates a distance x from the end of the source region 1 side of the channel region 3 to the direction toward the drain region 2 side and a vertical axis designates a channel potential V_(cs); and

FIG. 6 is a graph to show a relationship between a drain voltage V_(DS) and a drain current I_(DS) when a gate voltage is set at a constant value V_(DD), in which a horizontal axis designates the drain voltage V_(DS) and a vertical axis designates the drain current I_(DS).

DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT

The embodiment according to the present invention will be described in detail with reference to the accompanying drawings. FIG. 3 is a plan view to show the configuration of a MOSFET according to the embodiment of the present invention. As shown in FIG. 3, the MOSFET of the present embodiment is composed of a source region 1, a drain region 2, and a channel region 3 disposed between the source region 1 and the drain region 2. The width W(x) of the channel region 3 extends continuously from the end of the source region 1 side to the end of the drain region 2 side. The width of the channel region 3 follows the mathematical equation (1) described above. Further, the width of the source region 1 and the width of the drain region 2 are equal to the width of neighboring channel region 3. Still further, an insulating layer 4 is formed on the channel region 3 (at the front side of the paper in FIG. 3) and a gate 5 is formed on the insulating layer 4.

The mathematical equation (1) is a result obtained by determining a width function maximizing a drain current, by a variation method, under a constraint condition where the total amount of charges in the channel is kept constant when the drain voltage (V_(DS)) is a drain voltage (V_(dsat)) at which the drain current starts saturating in the conventional rectangular MOSFET.

In the mathematical equation (1), assuming that a carrier saturation velocity is V_(sat) (cm/sec) and a carrier mobility is μ (cm²/Vsec), the strength of carrier velocity saturation electric field E_(sat) (V/cm) is expressed by E_(sat)=2V_(sat)/μ. The carrier mobility (μ) can be measured. Also, the carrier saturation velocity (V_(sat)) can be estimated by the use of a full-band Monte Carlo simulation disclosed in “Monte Carlo analysis of electron transport in small semiconductor devices including band-structure and space-charge effects” authored by M. V. Fischetti and S. E. Laux, Physical Review B, VOL. 38, NO. 14, Page 9721-9745. Further, the physical meaning of a substrate charge effect coefficient (δ) is described in “Operation and modeling of the MOS transistor”, page 123-130, authored by Y. P. Tsividis. Still further, the standard width W₀ of the channel region is a width of a channel region of a conventional rectangular MOSFET in which, when V_(DS)=V_(dsat), the amount of all charges in the channel region is equal to that of the MOSFET according to the present invention.

A process of introducing the mathematical equation (1) will be described in the following. Assuming that a channel potential is V_(cs), a charge density is Q′, the gate capacitance per a unit area is C_(OX), and the amount of all charges in the channel region is A, the following equations are obtained. $I_{D} = {\mu \quad {W(x)}\left\{ {- {Q_{I}^{\prime}\left( V_{CS} \right)}} \right\} \frac{V_{CS}}{x}}$ Q_(I)^(′)(V_(CS)) = −C_(OX){V_(GS) − V_(T) − (1 + δ)V_(CS)}

The object of introducing the mathematical equation (1) is to determine W(x) maximizing the following value, ${{{W(0)}\left( {- {Q_{I}^{\prime}(0)}} \right)\frac{V_{CS}}{x}}}_{X = 0}$

while keeping the following A at a constant value. A = ∫₀^(L)W(x){−Q_(I)^(′)(V_(CS))}x_(V_(DS) = V_(dsat))

When the width W of the channel region is a constant value W₀, the following equation is obtained, $\begin{matrix} {A = {\int_{0}^{L}{{W(0)}\left( {- Q_{I}^{\prime}} \right){x}}}} \\ {= {\int_{0}^{V_{dsat}}{\frac{\mu \quad W_{0}^{2}}{I_{D0}}Q_{I}^{\prime 2}{V_{CS}}}}} \\ {= {\int_{0}^{V_{dsat}}{\frac{\mu \quad W_{0}^{2}}{I_{D0}}C_{OX}^{2}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}^{2}{V_{CS}}}}} \\ {= {\frac{\mu \quad W_{0}^{2}C_{OX}^{2}}{I_{D0}}\left\lbrack {\frac{- 1}{3\left( {1 + \delta} \right)}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}^{3}} \right\rbrack}_{0}^{V_{dsat}}} \\ {= {\frac{\mu \quad W_{0}^{2}C_{OX}^{2}}{I_{D0}} \times \frac{\left( {V_{GS} - V_{T}} \right)^{3}}{3\left( {1 + \delta} \right)}}} \end{matrix}$

where I_(DO) is given by the following equation. $\begin{matrix} {{I_{DO} = {\frac{W_{0}}{L}\mu \quad C_{OX}\left\{ {{\left( {V_{GS} - V_{T}} \right)V_{DS}} - {\frac{1}{2}\left( {1 + \delta} \right)V_{DS}^{2}}} \right\}}}}_{V_{DS} = V_{dsat}} \\ {= {\frac{W_{0}}{L}\mu \quad C_{OX}^{2}\frac{\left( {V_{GS} - V_{T}} \right)^{2}}{2\left( {1 + \delta} \right)}}} \end{matrix}$

Therefore, A is given by the following equation. $\begin{matrix} {A = {\frac{\mu \quad W_{0}^{2}{C_{OX}^{2}\left( {V_{GS} - V_{T}} \right)}^{3}}{3\left( {1 + \delta} \right)} \times \frac{L}{W_{0}\mu \quad C_{OX}} \times \frac{2\left( {1 + \delta} \right)}{\left( {V_{GS} - V_{T}} \right)^{2}}}} \\ {= {\frac{2}{3}{LW}_{0}{C_{OX}\left( {V_{GS} - V_{T}} \right)}}} \end{matrix}$

Assuming that a value of W(x)×(−Q₁′.(V_(CS))) is equal to a constant value B, A is obtained by following equation. $\begin{matrix} {A = {\int_{0}^{L}{B{x}}}} \\ {= {BL}} \end{matrix}$

Therefore, the following equations are obtained. $B = {\frac{2}{3}W_{0}{C_{OX}\left( {V_{GS} - V_{T}} \right)}}$ $\begin{matrix} {{W(x)} = \frac{B}{- {Q_{1}^{\prime}\left( V_{CS} \right)}}} \\ {= {\frac{2}{3}\frac{W_{0}{C_{OX}\left( {V_{GS} - V_{T}} \right)}}{C_{OX}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}}}} \end{matrix}$

From the result of the variation method, the following equations are obtained. $\frac{V_{CS}}{x} = {\frac{V_{dsat}}{L} = {const}}$ $V_{CS} = {\frac{x}{L}V_{dsat}}$

The following equations are obtained by these equations. $\begin{matrix} {{\begin{matrix} {I_{D} = {{W(0)}\left( {- {Q_{I}^{\prime}(0)}} \right)\mu \frac{V_{CS}}{x}}} \\ {= {\frac{2}{3}W_{0}{C_{OX}\left( {V_{GS} - V_{T}} \right)}\mu \frac{V_{dsat}}{L}}} \\ {= {\frac{W_{0}}{L}\mu \quad C_{OX}\quad \frac{\left( {V_{GS} - V_{T}} \right)^{2}}{\left( {1 + \delta} \right)} \times \frac{2}{3}}} \end{matrix}{I_{D} = {\frac{4}{3}I_{D0}}}{W(x)} = {\frac{2}{3}W_{0}\frac{\left( {V_{GS} - V_{T}} \right)}{\left( {V_{GS} - V_{T}} \right) - {\left( {1 + \delta} \right)\frac{x}{L}V_{dsat}}}}}{{W(x)} = {\frac{2}{3}W_{0}\frac{1}{\left( {1 - \frac{x}{L}} \right)}}}} & (2) \end{matrix}$

FIG. 4 is a graph to show a relationship between x and W(x) satisfying the mathematical equation (2) described above, in which a horizontal axis designates a distance x from the end of the source region 1 side of the channel region 3 to the direction toward the drain region 2 side and a vertical axis designates a width W(x) of the channel region 3 at the position of a distance x.

In order to eliminate the singularity at x=L, the velocity saturation type equation of BSIM is utilized. $\begin{matrix} {{A = {\int_{0}^{L}{{W_{0}\left( {- Q_{I}^{\prime}} \right)}{x}}}}{Q_{I}^{\prime} = {{- C_{OX}}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}}}{\frac{V_{CS}}{x} = \frac{I_{DO}}{{\mu \quad W_{0}C_{OX}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}} - \frac{I_{DO}}{E_{sat}}}}{{x} = {\left\lbrack {{\frac{\mu \quad W_{0}C_{OX}}{I_{D0}}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}} - \frac{1}{E_{sat}}} \right\rbrack {V_{CS}}}}\begin{matrix} {A = \quad {W_{0}{\int_{0}^{V_{dsat}}\left\lbrack {{\frac{\mu \quad W_{0}C_{OX}^{2}}{I_{D0}}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}^{2}} -} \right.}}} \\ {\left. \quad {\frac{C_{OX}}{E_{sat}}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}} \right\rbrack {V_{CS}}} \end{matrix}\begin{matrix} {A = \quad {W_{0}\left\lbrack {{{- \frac{\mu \quad W_{0}C_{OX}^{2}}{3\left( {1 + \delta} \right)I_{D0}}}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}^{3}} +} \right.}} \\ \left. \quad {\frac{C_{OX}}{2\left( {1 + \delta} \right)E_{sat}}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}^{2}} \right\rbrack_{0}^{V_{dsa}t} \end{matrix}\begin{matrix} {A = \quad {{\frac{\mu \quad W_{0}^{2}C_{OX}^{2}}{3\left( {1 + \delta} \right)I_{D0}}\left\lbrack {\left( {V_{GS} - V_{T}} \right)^{3} - \left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} \right\}^{3}} \right\rbrack} -}} \\ {\quad {\frac{W_{0}C_{OX}}{2\left( {1 + \delta} \right)E_{sat}}\left\lbrack {\left( {V_{GS} - V_{T}} \right)^{2} - \left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} \right\}^{2}} \right\rbrack}} \end{matrix}} & (3) \end{matrix}$

Also, I_(DO) is given by the following equation. $I_{D0} = {\mu \quad C_{OX}\frac{W_{0}}{L}\frac{1}{1 + \frac{V_{dsat}}{E_{sat}L}}\left\{ {V_{GS} - V_{T} - {\frac{\left( {1 + \delta} \right)}{2}V_{dsat}}} \right\} V_{dsat}}$

By substituting the I_(DO) into the mathematical equation (3), the following mathematical equation is obtained. $\begin{matrix} {\begin{matrix} {A = \quad {\frac{W_{0}^{2}C_{OX}^{2}}{3\left( {1 + \delta} \right)} \times \frac{L}{C_{OX}W_{0}}\left( {1 + \frac{V_{dsat}}{E_{sat}L}} \right) \times}} \\ {\quad {\frac{\left( {V_{CS} - V_{T}} \right)^{3} - \left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} \right\}^{3}}{\left\{ {V_{GS} - V_{T} - {\frac{1 + \delta}{2}V_{dsat}}} \right\} V_{dsat}} -}} \\ {\quad {\frac{W_{0}C_{OX}}{2\left( {1 + \delta} \right)E_{sat}}\left\lbrack {\left( {V_{GS} - V_{T}} \right)^{2} - \left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} \right\}^{2}} \right\rbrack}} \end{matrix}\begin{matrix} {A = \quad {\frac{W_{0}C_{OX}}{3\left( {1 + \delta} \right)}{L\left( {1 + \frac{V_{dsat}}{E_{sat}L}} \right)} \times}} \\ {\quad {\frac{\left( {V_{CS} - V_{T}} \right)^{3} - \left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} \right\}^{3}}{\left\{ {V_{GS} - V_{T} - {\frac{1 + \delta}{2}V_{dsat}}} \right\} V_{dsat}} -}} \\ {\quad {\frac{W_{0}C_{OX}}{2\left( {1 + \delta} \right)E_{sat}}\left\lbrack {\left( {V_{GS} - V_{T}} \right)^{2} - \left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} \right\}^{2}} \right\rbrack}} \end{matrix}} & (4) \end{matrix}$

Further, the following equation is assumed in BSIM. $V_{dsat} = \frac{E_{sat}{L\left( {V_{GS} - V_{T}} \right)}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}$

By the equation, this equation is changed as follows. $\begin{matrix} {{V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{dsat}}} = \quad {V_{GS} - V_{T} - \frac{\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \\ {= \quad \frac{\left( {V_{GS} - V_{T}} \right)^{2}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}} \end{matrix}$

Therefore, following equation is obtained. $\begin{matrix} {{V_{GS} - V_{T} - {\frac{\left( {1 + \delta} \right)}{2}V_{dsat}}} = {\frac{1}{2} \times \frac{{2\left( {V_{GS} - V_{T}} \right)^{2}} + {\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \\ {= {\frac{\left( {V_{GS} - V_{T}} \right) + {\frac{1}{2}\left( {1 + \delta} \right)E_{sat}L}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}\left( {V_{GS} - V_{T}} \right)}} \end{matrix}$

Therefore, following equation is obtained. $\begin{matrix} {{1 + \frac{V_{dsat}}{E_{sat}L}} = {1 + \frac{V_{GS} - V_{T}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \\ {= \frac{{2\left( {V_{GS} - V_{T}} \right)} + {\left( {1 + \delta} \right)E_{sat}L}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}} \end{matrix}$

When these mathematical equations are substituted into the mathematical equation (4), the following equation can be obtained. $\begin{matrix} {A = \quad {\frac{W_{0}C_{OX}}{3\left( {1 + \delta} \right)}L\frac{{2\left( {V_{GS} - V_{T}} \right)} + {\left( {1 + \delta} \right)E_{sat}L}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \times}} \\ {\quad {\frac{\frac{{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{3}\left( {V_{GS} - V_{T}} \right)^{3}} - \left( {V_{GS} - V_{T}} \right)^{6}}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{3}}}{\frac{\left( {V_{GS} - V_{T}} \right) + {\frac{1}{2}\left( {1 + \delta} \right)E_{sat}L}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}\left( {V_{GS} - V_{T}} \right)\frac{E_{sat}{L\left( {V_{GS} - V_{T}} \right)}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}} -}} \\ {\quad {\frac{W_{0}C_{OX}}{2\left( {1 + \delta} \right)E_{sat}L} \times \frac{{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}\left( {V_{GS} - V_{T}} \right)^{2}} - \left( {V_{GS} - V_{T}} \right)^{4}}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}}}} \\ {= \quad {{\frac{2W_{0}C_{OX}}{3\left( {1 + \delta} \right)E_{sat}} \times \frac{\left( {V_{GS} - V_{T}} \right)\left\lbrack {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{3} - \left( {V_{GS} - V_{T}} \right)^{3}} \right\rbrack}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GD} - V_{T}} \right\}^{2}}} -}} \\ {\quad {\frac{W_{0}C_{OX}}{2\left( {1 + \delta} \right){Esat}} \times \frac{\left( {V_{GS} - V_{T}} \right)^{2}\left( {1 + \delta} \right)E_{sat}L\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{GS} - V_{T}} \right)}} \right\}}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}}}} \\ {= \quad {\frac{{LW}_{0}C_{OX}}{6} \times \frac{\left( {V_{GS} - V_{T}} \right)}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} \times}} \\ {\quad \left\lbrack {{4\left\{ {\left( {1 + \delta} \right)E_{sat}L} \right\}^{2}} + {8\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}} + {4\left( {V_{GS} - V_{T}} \right)^{2}} +} \right.} \\ {\quad {{4\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}} + {4\left( {V_{GS} - V_{T}} \right)^{2}} + {4\left( {V_{GS} - V_{T}} \right)^{2}} -}} \\ \left. \quad {{3\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}} - {6\left( {V_{GS} - V_{T}} \right)^{2}}} \right\rbrack \\ {= \quad {\frac{{LW}_{0}C_{OX}}{6} \times \frac{\left( {V_{GS} - V_{T}} \right)}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} \times}} \\ {\quad \left\lbrack {4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2} \times \left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{GS} - V_{T}} \right)}} \right\} \left( {V_{GS} - V_{T}} \right)} \right\rbrack} \end{matrix}$

Assuming that a value of W(x)×(−Q₁′ (V_(CS))) is equal to a constant value B, the following equations are obtained. A = ∫₀^(L)Bx = BL $\begin{matrix} {B = \quad {\frac{W_{0}C_{OX}}{6} \times \frac{\left( {V_{GS} - V_{T}} \right)}{\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} \times}} \\ {\quad \left\lbrack {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} +} \right.} \\ \left. \quad {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{GS} - V_{T}} \right)}} \right\} \left( {V_{GS} - V_{T}} \right)} \right\rbrack \end{matrix}$ $\begin{matrix} {{W(x)} = \frac{B}{- {Q_{I}^{\prime}\left( V_{CS} \right)}}} \\ {= \frac{B}{C_{OX}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\}}} \end{matrix}$

On the other hand, since the following mathematical equation is established, I_(D) is determined as follows. $I_{D} = {{\mu \quad W_{0}C_{OX}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\} \frac{V_{CS}}{x}} - {\frac{1}{E_{sat}} \times \frac{V_{CS}}{x}I_{D}}}$ $I_{D} = \frac{\mu \quad W_{0}C_{OX}\left\{ {V_{GS} - V_{T} - {\left( {1 + \delta} \right)V_{CS}}} \right\} \frac{V_{CS}}{x}}{\left( {1 + {\frac{1}{Esat}\frac{V_{CS}}{x}}} \right)}$

According to a variational principle, the following equation is obtained. ${{\mu \frac{V_{CS}}{x}} - {\lambda \left( {1 + {\frac{1}{E_{sat}}\frac{V_{CS}}{x}}} \right)}} = 0$ ${\left( {\mu - \frac{\lambda}{E_{sat}}} \right)\frac{V_{CS}}{x}} = {{\lambda \therefore\frac{V_{CS}}{x}} = {const}}$ $V_{CS} = {{\frac{x}{L}V_{dsat}} = {\frac{E_{sat}\left( {V_{GS} - V_{T}} \right)}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}x}}$

Therefore, the following equation is obtained. ${W(x)} = \frac{B}{C_{OX}\left\{ {V_{GS} - V_{T} - {\frac{\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \times \frac{x}{L}}} \right\}}$ $\begin{matrix} {{W(x)} = \quad \frac{B\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}}{C_{OX}\left\{ {{\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}\left( {1 - \frac{x}{L}} \right)} + \left( {V_{GS} - V_{T}} \right)^{2}} \right\}}} \\ {= \quad {\frac{W_{0}}{6} \times \frac{V_{GS} - V_{T}}{{\left( {1 + \delta} \right)E_{sat}{L\left( {V_{GS} - V_{T}} \right)}\left( {1 - \frac{x}{L}} \right)} + \left( {V_{GS} - V_{T}} \right)^{2}} \times}} \\ {\quad \frac{\begin{matrix} \left\{ {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} +} \right. \\ {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{GS} - V_{T}} \right)}} \right\} \left( {V_{GS} - V_{T}} \right)} \end{matrix}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}} \end{matrix}$

By further changing this equation, the following mathematical equation (1) is derived. $\begin{matrix} {{W(x)} = {\frac{\begin{matrix} {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}^{2}} +} \\ {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{DD} - V_{T}} \right)}} \right\} \left( {V_{DD} - V_{T}} \right)} \end{matrix}}{\begin{matrix} {6\left\{ {{\left( {1 + \delta} \right)E_{sat}{L\left( {1 - \frac{x}{L}} \right)}} + V_{DD} - V_{T}} \right\}} \\ \left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\} \end{matrix}}\quad W_{O}}} & (1) \end{matrix}$

For example, assuming that an n-type MOSFET has a channel region length L of 0.05 μm, V_(DD) is 1.0V, V_(T) is 0.2V, δ is 0.25, μ is 300 cm²/Vsec, and V_(sat) is 2×10⁷ cm/sec in consideration of the velocity overshoot effect of the electron, these values are substituted into the mathematical equation (1). As a result, the mathematical equation (1) is expressed by the mathematical equation (5). $\begin{matrix} {{W(x)} = {\frac{12.62}{{8.17\left( {1 - \frac{x\lbrack{cm}\rbrack}{5 \times 10^{- 6}}} \right)} + 7.84}W_{0}}} & (5) \end{matrix}$

Next, the operation of the present embodiment will be described. If the width of the channel region is set such that it satisfies the mathematical equation (1), when V_(DS)=V_(dsat), the amount of all the charges in the inversion layer integrated in the direction of channel width at a position x in the channel region is not dependent on x but is kept at a constant value.

FIG. 5 is a graph to show a channel potential distribution of the MOSFET according to the present embodiment, in which a horizontal axis designates a distance x from the end of the source region 1 side of the channel region 3 to the direction toward the drain region 2 side and a vertical axis designates a channel potential V_(cs). FIG. 5 is a graph to show a channel potential V_(CS) (x) when V_(DS)=V_(dsat). As shown in FIG. 5, in the present embodiment, when V_(DS)=V_(dsat), the channel potential V_(CS) (x) changes linearly along the direction of the channel. For this reason, a stronger electric field is applied in the direction of the channel at the end of the source side of the channel region.

On the other hand, when V_(DS)=V_(dsat), the width of the channel region in the conventional rectangular MOSFET in which the amount of all the charges in the channel region is equal to that of the MOSFET of the present embodiment is W₀ (see FIG. 1). In this MOSFET, the width of the channel region does not change according to the mathematical equation (1) and hence, as shown in FIG. 2, the gradient of the channel potential V_(CS) decreases in the vicinity of the source region.

Next, the effect of the present embodiment will be described. FIG. 6 is a graph to show a relationship between a drain voltage V_(DS) and a drain current I_(DS) when a gate voltage is set at a constant value V_(DD), in which a horizontal axis designates the drain voltage V_(DS) and a vertical axis designates the drain current I_(DS). In FIG. 6, a line 6 shows the relationship between the drain voltage V_(DS) and the drain current I_(DS) in the MOSFET of the present embodiment, and a line 7 shows the relationship between the drain voltage V_(DS) and the drain current I_(DS) in a conventional MOSFET in which the width of the channel region is a constant value W₀. As shown in FIG. 6, the drain current I_(DS) (shown by the line 6) in the MOSFET of the present embodiment becomes larger than the drain current I_(DS) (shown by the line 7) in the conventional MOSFET in which the width of the channel region is a constant value W₀. When V_(DS)=V_(dsat), the ration (I_(D)/I_(D0)) between them is expressed by the following mathematical equation (6).

From the results of calculation described above, the following equation is obtained. $\begin{matrix} {I_{D} = {{W(0)}\left( {- {Q_{I}^{\prime}(0)}} \right)\mu \frac{V_{CS}}{x}}} \\ {= {\frac{B}{C_{OX}\left( {V_{GS} - V_{T}} \right)} \times \left( {V_{GS} - V_{T}} \right)\mu \frac{E_{sat}\left( {V_{GS} - V_{T}} \right)}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \\ {= {B\quad \mu \frac{E_{sat}\left( {V_{GS} - V_{T}} \right)}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \end{matrix}$

On the other hand, the following equation is obtained. $\begin{matrix} {I_{D0} = \quad {\mu \quad C_{OX}\frac{W_{0}}{L} \times \frac{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}{{2\left( {V_{GS} - V_{T}} \right)} + {\left( {1 + \delta} \right)E_{sat}L}} \times}} \\ {\quad {\frac{\left( {V_{GS} - V_{T}} \right) + {\frac{1}{2}\left( {1 + \delta} \right)E_{sat}L}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}\left( {V_{GS} - V_{T}} \right)\frac{E_{sat}{L\left( {V_{GS} - V_{T}} \right)}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \\ {= \quad {\mu \quad C_{OX}\frac{W_{0}}{L} \times \frac{1}{2} \times \frac{E_{sat}{L\left( {V_{GS} - V_{T}} \right)}^{2}}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}}} \end{matrix}$ $\begin{matrix} {{\therefore\frac{I_{D}}{I_{D0}}} = \quad {B\quad \mu \frac{E_{sat}\left( {V_{GS} - V_{T}} \right)}{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \times \frac{2L}{\mu \quad C_{OX}W_{0}} \times \frac{{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}}{E_{sat}{L\left( {V_{GS} - V_{T}} \right)}^{2}}}} \\ {= \quad \frac{2B}{W_{0}{C_{OX}\left( {V_{GS} - V_{T}} \right)}}} \\ {= \quad {2\frac{W_{0}C_{OX}}{6} \times \frac{1}{W_{0}C_{OX}\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} \times}} \\ {\quad \left\lbrack {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{GS} - V_{T}} \right\}^{2}} +} \right.} \\ \left. \quad {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{GS} - V_{T}} \right)}} \right\} \left( {V_{GS} - V_{T}} \right)} \right\rbrack \end{matrix}$

By further changing this mathematical equation, the following mathematical equation (6) is obtained at V_(GS)=V_(DD). $\begin{matrix} {\frac{I_{D}}{I_{D0}} = \frac{\begin{matrix} \left\lbrack {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}^{2}} +} \right. \\ {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{DD} - V_{T}} \right)}} \right\} \left( {V_{DD} - V_{T}} \right)} \end{matrix}}{3\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}^{2}}} & (6) \end{matrix}$

If the conditions under which the mathematical equation (5) is derived, that is, L=0.05 μm, V_(DD)=1.0 V, V_(T)=0.2 V, δ=0.25, μ=300 cm²/Vsec, and V_(sat)=2×10⁷ cm/sec, are substituted into the mathematical equation (6), then the value of (I_(D)/I_(D0)) becomes 1.58. Therefore, the MOSFET of the present invention can increase the drain current in comparison with the conventional MOSFET in which the amount of all the charges in the channel region, when V_(DS)=V_(dsat), is equal to that of the MOSFET of the present invention.

On the other hand, assuming that the gate capacitance per a unit area is C_(OX), the amount of all the charges Q_(CH) in the inversion layer in the channel region when V_(DS)=V_(dsat) is expressed by the following mathematical equation (7). As shown in the following mathematical equation (7), the amount of all the charges in the inversion layer in the MOSFET of the present invention is equal to the amount of all the charges in the inversion layer in the conventional MOSFET in which the width of the channel region is W_(O). Therefore, the present invention can increase the drain current while keeping the channel capacitance when V_(DS)=V_(dsat) at the same value as the conventional MOSFET. $\begin{matrix} \begin{matrix} {Q_{CH} = {\int_{0}^{L}{W_{0}C_{OX}\left\{ {V_{DD} - V_{T} - {\left( {1 + \delta} \right){V_{CS0}(c)}}} \right\} {x}}}} \\ {= {\int_{0}^{L}{W_{0}C_{OX}\left\{ {V_{DD} - V_{T} - {\left( {1 + \delta} \right){V_{CS}(x)}}} \right\} {x}}}} \end{matrix} & (7) \end{matrix}$

As described above, according to the present invention, by suitably selecting the width of the channel region according to distances from the source region and the drain region, the drain current of the MOSFET can remarkably be increased. 

What is claimed is:
 1. A hyperbolic type channel MOSFET comprising: a source region; a drain region; and a channel region disposed between said source region and said drain region; in said channel region, the width of said channel region changing such that the width W(x) of the source region is smaller than the width W(x) of the drain region for all values of x, wherein a distance from the end of the source region side to the direction toward the drain region side is x, and the width of said channel region at the position of the distance x is W(x).
 2. The hyperbolic type channel MOSFET according to claim 1, wherein the width of said source region is equal to or larger than the neighboring channel region.
 3. The hyperbolic type channel MOSFET according to claim 1, wherein the width of said drain region is equal to or larger than the neighboring channel region.
 4. The hyperbolic type channel MOSFET according to claim 1, further comprising: an insulating film formed on said channel region; and a gate electrode formed on said insulating film.
 5. The hyperbolic type channel MOSFET according to claim 1, wherein the width of said channel region changes according to the following mathematical equation: ${W(x)} = {\frac{\begin{matrix} {{4\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}^{2}} +} \\ {\left\{ {{\left( {1 + \delta} \right)E_{sat}L} + {2\left( {V_{DD} - V_{T}} \right)}} \right\} \left( {V_{DD} - V_{T}} \right)} \end{matrix}}{6\left\{ {{\left( {1 + \delta} \right)E_{sat}{L\left( {1 - {x/L}} \right)}} + V_{DD} - V_{T}} \right\} \left\{ {{\left( {1 + \delta} \right)E_{sat}L} + V_{DD} - V_{T}} \right\}}W_{o}}$

wherein said distance from the end of the source region side to the direction toward the drain region side is x, said width of said channel region at the position of the distance x is W(x), a substrate charge effect coefficient is δ, the length from the end of the source region side to the end of the drain side is L, a gate voltage is V_(DD), a threshold voltage is V_(T), a standard width of the channel region is W₀, and a strength of a carrier velocity saturation electric field is E_(sat). 